FE Environmental Surface Water Resources and Hydrology Practice Problems

The current FE Environmental specification assigns 9-14 of the 110 exam questions to Surface Water Resources and Hydrology. Form A contains 10 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

Three recording rain gauges serve a 45.0 km $^{2}$ basin. Storm depths and the corresponding Thiessen polygon areas are: gauge 1, 62 mm over 14.0 km $^{2}$ ; gauge 2, 88 mm over 22.5 km $^{2}$ ; gauge 3, 45 mm over 8.5 km $^{2}$ . The average storm depth over the basin is most nearly:

(A)58.4 mm
(B)62.0 mm
(C)65.0 mm
(D)71.8 mm

Show worked solution

Answer: (D)

Weighting each gauge by its polygon area gives 71.8 mm, well above the unweighted mean because the wettest station governs half the basin.

Area weighting governs here because the polygons differ by nearly a factor of three, and an unweighted mean would give the driest station equal standing.

\overset{ˉ}{P} = \frac{\sum P _{i} A _{i}}{\sum A _{i}}

Gauge 2 holds half the basin area and therefore controls the accumulated product.

\sum P_{i} A_{i} = 868 + 1980 + 382.5 = 3230.5 km^{2} \cdot mm

The equivalent uniform depth sits above the 65 mm arithmetic mean precisely because the wettest gauge carries the largest polygon.

\overset{ˉ}{P} = \frac{3230.5}{45.0} = 71.8 mm

FE Reference Handbook — Water Resources and Environmental Engineering: Precipitation, Areal Average by Thiessen Polygon Weighting

Why the other choices appear

(A)58.4 mm comes from inverting the weights, using 1/A_i instead of A_i: 13.634 / 0.23352 = 58.4 mm.
(B)62.0 mm is the single nearest-station depth adopted for the whole basin instead of weighting all three gauges.
(C)65.0 mm is the arithmetic mean of the three depths, (62 + 88 + 45)/3, with the polygon areas ignored.